Posted by Jay on Tuesday, April 8, 2008 at 8:57pm.
The following answers the question but it doesn't involve percent ionization unless you want to calculate it from the Ka values in the table.
HF ==> H^+ + F^-
Use the ICE table.
Change in conen and
(HF) = 0.1 M
(H^+) = 0
(F^-) = 0
Change in concn:
(H^+) = +x
(F^-) = +x
(HF) = -x
(HF) = 0.1-x
(H^+) = 0 + x = x
(F^-) = 0 + x = x
Write the Ka expression. For HF it is
(H^+)(F^-)/(HF) = Ka.
Look up Ka in your tables, plug in the equilibrium concn from above, and solve for x. That is the (H^+).
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